$ B = \left[\begin{array}{rr}5 & 3 \\ 5 & 3\end{array}\right]$ $ A = \left[\begin{array}{rrr}4 & 3 & -1 \\ -1 & 0 & -2\end{array}\right]$ What is $ B A$ ?
Explanation: Because $ B$ has dimensions $(2\times2)$ and $ A$ has dimensions $(2\times3)$ , the answer matrix will have dimensions $(2\times3)$ $ B A = \left[\begin{array}{rr}{5} & {3} \\ {5} & {3}\end{array}\right] \left[\begin{array}{rrr}{4} & \color{#DF0030}{3} & \color{#9D38BD}{-1} \\ {-1} & \color{#DF0030}{0} & \color{#9D38BD}{-2}\end{array}\right] = \left[\begin{array}{rrr}? & ? & ? \\ ? & ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rrr}{5}\cdot{4}+{3}\cdot{-1} & ? & ? \\ ? & ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rrr}{5}\cdot{4}+{3}\cdot{-1} & ? & ? \\ {5}\cdot{4}+{3}\cdot{-1} & ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rrr}{5}\cdot{4}+{3}\cdot{-1} & {5}\cdot\color{#DF0030}{3}+{3}\cdot\color{#DF0030}{0} & ? \\ {5}\cdot{4}+{3}\cdot{-1} & ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rrr}{5}\cdot{4}+{3}\cdot{-1} & {5}\cdot\color{#DF0030}{3}+{3}\cdot\color{#DF0030}{0} & {5}\cdot\color{#9D38BD}{-1}+{3}\cdot\color{#9D38BD}{-2} \\ {5}\cdot{4}+{3}\cdot{-1} & {5}\cdot\color{#DF0030}{3}+{3}\cdot\color{#DF0030}{0} & {5}\cdot\color{#9D38BD}{-1}+{3}\cdot\color{#9D38BD}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rrr}17 & 15 & -11 \\ 17 & 15 & -11\end{array}\right] $